$f(x,y) = e^x - x^2$ What is $\dfrac{\partial f}{\partial y}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $e^x - 2x$ (Choice B) B $e^x - x$ (Choice C) C $-2x$ (Choice D) D $0$
Answer: We want to find $\dfrac{\partial f}{\partial y}$, which is the partial derivative of $f$ with respect to $y$. When we take a partial derivative with respect to $y$, we treat $x$ as if it were a constant. Let's break $f(x, y)$ down term by term. $\begin{aligned} &\dfrac{\partial}{\partial y} \left[ e^x \right] = 0 \\ \\ &\dfrac{\partial}{\partial y} \left[ -2x \right] = 0 \end{aligned}$ Adding the terms back together, we get the partial derivative. In conclusion: $\dfrac{\partial f}{\partial y} = 0 + 0 = 0$